Hdu 2888 Check Corners (二维RMQ (ST))

题目链接：

　　Hdu 2888 Check Corners

题目描述：

　　给出一个n*m的矩阵，问以(r1,c1)为左上角，(r2,c2)为右下角的子矩阵中最大的元素值是否为子矩阵的顶点？

解题思路：

　　二维区间最值查询，可以用二维的ST算法，dp[x][y][i][j]表示x轴上[x,x+(1<<i)-1]与y轴上[y-(1<<j)+1,y]组成的子矩阵中的最值。预处理的时候处理出来子矩阵的最值，查询的时候对于x,y轴上的查询区间[m, n],都要找到一个k,k满足 n-m+1 < 2^^(k+1)，假设x轴上为k1，y轴上为k2，然后把x轴上[r1,r2]分为[r1, r1+(1<<k1)-1]与[r2-(1<<k1)+1, r2],y轴类似。根据对x，y轴的划分把查询子矩阵分成4部分，然后选取4部分中的最值即可。

 

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

const int maxn = 305;
int dmax[maxn][maxn][9][9];
int arr[maxn][maxn], LOG[maxn];

void init_RMQ (int n, int m)
{
    LOG[0] = -1;
    for (int i=1; i<maxn; i++)
        LOG[i] = i&(i-1)?LOG[i-1]:LOG[i-1]+1;

    for (int i=1; i<=n; i++)
        for (int j=1; j<=m; j++)
            dmax[i][j][0][0] = arr[i][j];

    for (int i=0; i<=LOG[n]; i++)
        for (int j=0; j<=LOG[m]; j++)
        {
            if (i==0 && j==0)   continue;
            
            for (int row=1; row+(1<<i)-1<=n; row++)
                for (int col=1; col+(1<<j)-1<=m; col++)
                    if (i == 0)
                        dmax[row][col][i][j] = max (dmax[row][col][i][j-1], dmax[row][col+(1<<(j-1))][i][j-1]);
                    else
                        dmax[row][col][i][j] = max (dmax[row][col][i-1][j], dmax[row+(1<<(i-1))][col][i-1][j]);
        }
}
int ST (int x1, int y1, int x2, int y2)
{//(x1,y1)为左上角，(x2,y2)为右下角
    int k1 = LOG [x2 - x1 + 1];
    int k2 = LOG [y2 - y1 + 1];
    int mm1 = dmax [x1][y1][k1][k2];
    int mm2 = dmax [x1][y2-(1<<k2)+1][k1][k2];
    int mm3 = dmax [x2-(1<<k1)+1][y1][k1][k2];
    int mm4 = dmax [x2-(1<<k1)+1][y2-(1<<k2)+1][k1][k2];
    return max (max (mm1, mm2), max (mm3, mm4));
}

int main ()
{
    int n, m;
    while (scanf ("%d %d", &n, &m) != EOF)
    {
        for (int i=1; i<=n; i++)
            for (int j=1; j<=m; j++)
                scanf ("%d", &arr[i][j]);

        init_RMQ(n, m);

        int x1, y1, x2, y2, q, ans;
        scanf ("%d", &q);
        while (q --)
        {
            scanf ("%d %d %d %d", &x1, &y1, &x2, &y2);
            ans = ST (x1, y1, x2, y2);
            printf ("%d ", ans);

            if (arr[x1][y1]==ans||arr[x1][y2]==ans||arr[x2][y1]==ans||arr[x2][y2]==ans)
                printf ("yes\n");
            else
                printf ("no\n");

        }
    }
    return 0;
}